If a graph G is H-decomposable does it imply that G has H-factor?
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No. An $H$-factor is a set of vertex-disjoint copies of $H$ that partition the vertex set of a graph, while an $H$-decomposition is a set of edge-disjoint copies of $H$ that partition the edge set of a graph.
For example, let $H = K_3$ and let $G$ be the $5$-vertex graph consisting of two triangles with a point in common. Then $G$ clearly has an $H$-decomposition, but, since $3 \nmid 5$, it cannot have an $H$-factor.